Writing…

.vscode/ltex.dictionary.en-GB.txt

@@ -24,3 +24,5 @@ Patakfalvi
 Zsolt
 Kebekus
 Albanese
+Hirzebruch
+multiplicitity

01-intro.tex

@@ -8,15 +8,15 @@
 \section{The $\cC$-Albanese morphism in the presence of rational curves}
 \subversionInfo
 
+\begin{setting}\label{set:1}
+  Let $X$ be a compact Kähler manifold and let $x \in X$ be any point. Assume
+  that an Albanese of the $\cC$-pair $(X,0)$ exists.
+\end{setting}  
+
 \begin{thm}\label{thm:1}%
-  Let $X$ be a projective manifold and let $x \in X$ be any point. If $C \subset
+  In Setting~\ref{set:1}, let $C \subset X$ be a rational curve.  Then, the
-  X$ is a rational curve, then the Albanese morphism\watchOut{Stefan 03Jun24:
+  Albanese morphism $\alb_x(X,0) : X \to \Alb_x(X,0)$ maps the curve $C$ to a
-  Need to make assumptions to ensure that the Albanese exists.} of the
+  point.
-  $\cC$-pair $(X,0)$,
-  \[
-    \alb_x(X,0) : X \to \Alb_x(X,0),
-  \]
-  maps the curve $C$ to a point.
 \end{thm}
 \begin{proof}
   The normalization of $C$ yields a diagram 
@@ -49,6 +49,53 @@
   The claim follows immediately once we observe that $\Alb_0(\bP¹,0)$ is a point.
 \end{proof}
 
+\begin{cor}\label{cor:3}%
+  In Setting~\ref{set:1}, assume that $X$ is rationally connected.  Then,
+  $\Alb_x(X,0)$ is a point.
+\end{cor}
+
+\begin{example}[Theorem~\ref{thm:1} is wrong for singular spaces]
+  Let $\pi : S \to \bP¹$ be one of the rational ruled ``Hirzebruch'' surfaces.
+  Let $C_S \subset S$ be any section.  Construct a commutative diagram as
+  follows,
+  \[
+    \begin{tikzcd}[column sep=2cm]
+      & S_2 \ar[r, "\alpha\text{, blow-up}"] \ar[d, "\gamma\text{, contraction}"'] & S_1 \ar[r, "\beta\text{, blow-up}"] & S \ar[r, "\pi\text{, fibre bundle}"] & \bP¹ \ar[d, equal] \\
+      C \ar[r, "\iota"'] & X \ar[rrr, "\rho\text{, rational fibration}"'] & & & \bP¹.
+    \end{tikzcd}
+  \]
+  \begin{itemize}
+    \item Choose four distinct points $x_1, …, x_4 \in \bP¹$.
+    \item Choose four points $s_\bullet \in \pi^{-1}(x_\bullet) \in \bP¹$.
+    \item Let $\beta$ be the blow-up up of the four points $s_\bullet$.  
+    \item The surface $S_1$ is smooth.  The fibres $F_{1\bullet} :=
+    (\pi\circ\beta)^{-1}(x_\bullet)$ are reduced. Each fibre $F_{1\bullet}$
+    consists of two $(-1)$-curves, meeting transversally in a point
+    $s_{1\bullet}$.
+    \item Let $\alpha$ be the blow-up up of the four points $s_{1\bullet}$.  
+    \item The surface $S_2$ is smooth but the fibres $F_{2\bullet} :=
+    (\pi\circ\beta\circ\alpha)^{-1}(x_\bullet)$ are no longer reduced. Each
+    fibre $F_{2\bullet}$ consists of two reduced $(-1)$-curves and one
+    $(-2)$-curve $F'_{2\bullet}$ of multiplicity two.
+    \item Let $\gamma$ be the contraction of the four points disjoint
+    $(-2)$-curves $F'_{2\bullet}$.  The map $\pi\circ\beta\circ\alpha$  factors
+    via the contraction map because we contract fibre components only.
+    \item Let $C \subset X$ be the strict transform of the section $C_S$.
+  \end{itemize}
+  The surface $X$ is then singular, with four quotient singularities of type
+  $A_1$ over the $x_\bullet$. All fibres of $\rho$ are supported on smooth
+  rational curves, but the fibres over $x_\bullet$ have multiplicitity two and
+  pass through the singularities.
+
+  The criterion for $\cC$-morphism spelled out in \cite{orbiAlb1} quickly
+  implies that $\rho$ is a $\cC$-morphism between the pair $(X,0)$ and the torus
+  quotient $(\bP¹, \frac{1}{2}·\sum_i x_i)$.  The universal property of the
+  Albanese immediately implies that the map $\rho$  factors via $\alb_x(X,0)$. A
+  more detailed analysis, applying Theorem~\ref{thm:1} to the smooth fibres of
+  $\rho$, shows that the torus quotient $(\bP¹, \frac{1}{2}·\sum_i x_i)$ is
+  equal to the Albanese and that $\rho$ is the Albanese map.
+\end{example}
+
 \begin{itemize}
   \item \todo{Need example where Theorem~\ref{thm:1} fails if $X$ is singular.}
 \end{itemize}